Dynamics of Uniform Circular Motion?

Question

At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders “plastered” against the wall. Suppose the radius of the room is 3.30 m and the speed of the wall is 10.0 m/s when the floor falls away. (a) How much centripetal force acts on a 55.0-kg rider? (b) What is the minimum coefficient of static friction that must exist between a rider’s back and the wall, if the rider is to remain in place when the floor drops away?
(a) Fc=mv^2/r =(55 kg x (10.0 m/s)^2)/ 3.30m = 1666.67 N

(b) Fc=FMax= usFN = usmg
usmg=mv^2/r
us=(((55kg) (10)^2)/ (3.3))/ (55) (9.8) = 3.09

I really need help on part b. Thanks in advance.

Answer 1

Your equation is upside down:

mu*Fc = mg to keep someone from slipping vertically

Answer 2

The radial acceleration a = v^2/r; the position v is the consistent tangential % and r is the radius of the Ferris Wheel. a is continually pointing outward alongside the spokes and remote from axis of rotation on the hub of the wheel. The acceleration by way of gravity is g; that is continually downward in the route of the floor. So even as the rider is on the bottom, a is pointing instantly down remote from the hub and alined with g, that is continually down. in which case, the internet acceleration is a' = g + a. even as on excellent, a is now pointing instantly up remote from the hub and counter course to g, that is continually down. in which case, the internet acceleration is a" = g - a. for this reason the internet, known rigidity even as the rider is on the bottom of his turn is F = ma' = m(g + a) >= mg = W the position W is the rider's weight. for this reason F >= W on the bottom. And the internet, known rigidity even as the rider is on excellent of the turn is F = ma" = m(g - a) <= mg = W. for this reason F <= W even as the rider is on the total. note that the = is valid if and on condition that a = 0, meaning v = 0 so the Ferris Wheel isn't transferring.